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Q. 6.35

Expert-verifiedFound in: Page 246

Book edition
1st

Author(s)
Daniel V. Schroeder

Pages
356 pages

ISBN
9780201380279

Verify from Maxwell speed distribution that the most likely speed of a molecule is $\sqrt{\raisebox{1ex}{$2kT$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}$.

The most likely speed of a molecule is ${v}_{\mathrm{mp}}=\sqrt{\raisebox{1ex}{$2kT$}\!\left/ \!\raisebox{-1ex}{$m$}\right.}$.

Maxwell speed distribution function is given by

$D\left(v\right)={\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4\pi {v}^{2}{e}^{-\raisebox{1ex}{$m{v}^{2}$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}..........................\left(1\right)$

Differentiate both sides of equation (1) with respect to $v$.

role="math" localid="1646971259346" $\begin{array}{rcl}\frac{dD\left(v\right)}{dv}& =& \frac{d}{dv}\left[{\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4\pi {v}^{2}{e}^{-\raisebox{1ex}{$m{v}^{2}$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}\right]\\ & =& {\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}4\pi \left[2v{e}^{-\raisebox{1ex}{$m{v}^{2}$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}-\frac{2m{v}^{3}}{2kT}{e}^{-\raisebox{1ex}{$m{v}^{2}$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}\right]\\ & =& 8\pi {\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}v{e}^{-\raisebox{1ex}{$m{v}^{2}$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}\left[1-\frac{m{v}^{2}}{2kT}\right]............................\left(2\right)\end{array}$

At most likely speed of the gas molecules, the derivative of the Maxwell speed distribution function will be zero.

Substitute ${v}_{\mathrm{mp}}$ for $v$ and $0$ for $\begin{array}{rcl}& & \frac{dD\left(v\right)}{dv}\end{array}$ into equation (2) and simplify to obtain the most likely speed.

$\begin{array}{rcl}0& =& 8\pi {\left(\frac{m}{2\pi kT}\right)}^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}v{e}^{-\raisebox{1ex}{$m{v}^{2}$}\!\left/ \!\raisebox{-1ex}{$2kT$}\right.}\left[1-\frac{m{v}^{2}}{2kT}\right]\\ \left[1-\frac{m{{v}_{\mathrm{mp}}}^{2}}{2kT}\right]& =& 0[\mathrm{As}\mathrm{v}\ne 0]\\ {v}_{\mathrm{mp}}& =& \sqrt{\frac{2kT}{m}}\end{array}$

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